Question

已知正项数列｛a｝满足a1=1/2,且a(n+1)=an/(1+an) 1, 求正项数列｛a｝的通项公式 2,求和:a1/1+a2/2+......

要过程
2,求和:a1/1 + a2/2 +...... +an/n

Answer 1

n>=3,
a(n)=a(n-1)/(1+a(n-1))=(a(n-2)/(1+a(n-2)))/(1+a(n-2)/(1+a(n-2)))
=a(n-2)/(1+2a(n-2))=(a(n-3)/(1+a(n-3)))/(1+2a(n-3)/(1+a(n-3)))
=a(n-3)/(1+3a(n-3))=.......
.....
=a2/(1+(n-2)a2)
=a1/(1+(n-1)a1)
=1/(n+1)
n=1,2通项满足通项公式，a(n)=1/(n+1)
a(n)/n=1/(n(n+1))=1/n-1/(n+1)
Sn=1-1/2+1/2-1/3+.....+1/n-1/(n+1)
=1-1/(n+1) 无数项相加时，Sn=1

Answer 2

取倒数
1/a(n+1)=(1+an)/an=1/an+1
所以1/an等差，d=1
则1/an=1/a1+(n-1)d=n+1
an=1/(n+1)

a1/1+a2/2+......这个写错了吧

Answer 3

an=1/(n+1)